Bifalgorithm post · Computer science one · String Library

How does strcat append one C string to another?

How does strcat append one C string to another? (Computer science one · String Library) — "A concatenation function is provided that allows you to append one s…

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The strcat() function appends one string to the end of another.
Function signature: char *strcat(char *dest, const char *src);
The source string is appended to the destination string.
Like strcpy(), strcat() operates on C strings.
The destination string must be large enough for the appended source string.
An insufficient destination buffer can cause undefined behavior.
strcat() copies the null-terminating character, producing a valid null-terminated string.
Example: constructing the formatted name "Waits, Tom".
Allocate space for formattedName with malloc().
Copy lastName into formattedName using strcpy().
Append ", " to formattedName using strcat().
Append firstName to formattedName using strcat().
The resulting formattedName contains "Waits, Tom".

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How does strcat append one C string to another?

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Computer science one

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sam23

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"A concatenation function is provided that allows you to append one string to the end of another.\n\nchar *strcat(char *dest, const char *src);\n\nSimilar to the strcpy() function, strcat() appends the “source” (src) string to the end of the “destination” (dest) string. It is your responsibility to ensure that the destination string is large enough to accommodate the source string. If it is not, then it could lead to undefined behavior as strcat() overwrites memory after the end of the destination string. Further, strcat() will copy the null-terminating character for you so that the resulting string (now stored in dest) is a valid null-terminated string.\n\n1 char *formattedName = (char *)malloc(11 * sizeof(char));\n\n2 //copy \"Waits\" into formattedName\n\n3 strcpy(formattedName, lastName);\n\n4 //append a \", \" to formattedName\n\n5 strcat(formattedName, \", \");\n\n6 //append the firstName to the formattedName\n\n7 strcat(formattedName, firstName);\n\n8 //formattedName now contains \"Waits, Tom\""

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